By Dr J Badshah University ProfessorcumChief scientist Dairy Engineering Department SGIDT PATNA Basu Patna Pipe Discharging from a Reservoir Applying Bernoullis Theorems between A and ID: 1019350
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1. Flow through Pipes discharging from a ReservoirByDr. J. BadshahUniversity Professor-cum-Chief scientistDairy Engineering DepartmentSGIDT, PATNA(Basu, Patna)
2. Pipe Discharging from a ReservoirApplying Bernoullis Theorems between A and B hL v2/2g E.G.L. hf H H.G.L. A B
3. Pipe Discharging from a reservoirApplying Bernoullis Theorems equation to points A ( at the joint with reservoir ) and B (At the end of pipe):H = hL+ hf + V2/2g, hL = Loss of head at the entrancehf = Loss due to friction in the entire lengthV2/2g = Distance between hydraulic gradient line and energy Gradient lineIf the entrance to the pipe is flush with the reservoir, hL = 0.5 V2/2g and
4. Pipe Discharging from a reservoirFrictional head is given by, hf = f L/D (V2/2g )H = V2/2g [1.50 + f L/D]For Long pipes, the term fL/D is very large compared to 1.50, then in such cases, the loss of head at the entrance and exit may be neglected. When the length of pipe is greater than 1000 D, only the friction loss need to be considered.Numericals will be followed on white board by sharing it.
5. Numericals on Pipe Discharging from a reservoirWater is discharged from a large reservoir to atmosphere through a 10 cm diameter and 500 m long pipe. Find the discharge if the outlet is 15 m below the free surface of water in the reservoir. Assume the entry to the pipe as sharp. Take friction factor f = 0.04.Solution: H = V2/2g [1.50 + f L/D]15 = V2/2g [1.50 + 0.04. 500/ 0.10]Therefore V = 1.21 m/sAnd Discharge Q = AV = π/4 x 0.1x0.1x1.21 = 0.0095 cumecs
6. Pipe Connecting Two ReservoirsThe liquid flows from the higher reservoir to the lower reservoir. The total losses is equal to the difference of liquid levels in the two reservoir. If the difference of liquid levels in the two reservoirs = HH = hL+ hf + V2/2g = 0.5 V2/2g + hf + V2/2g hL = Loss of head at the entrancehf = Loss due to friction in the entire lengthV2/2g = Distance between hydraulic gradient line and energy Gradient line
7. Pipes in series connecting between two reservoirsDifference in liquid levels between two reservoirs = hL1 + hf1 + hL2 + hf2 + hL3 + hf3 + V3 2 /2g, wherehL1 = Losses at entrancehL2 = Losses at contraction in figurehL3 = Losses at enlargementHf1 , hf2, hf3 are losses due to friction in three pipes in seriesV3 = Velocity in pipe 3Q1 = Q2 = Q3 and A1V1 = A2V2= A3V3
8. ..Thank You